## Calculus: Early Transcendentals (2nd Edition)

Published by Pearson

# Chapter 7 - Integration Techniques - 7.4 Trigonometric Substitutions - 7.4 Exercises - Page 537: 34

#### Answer

$$2\ln \left| {2x + \sqrt {4{x^2} - 1} } \right| - \frac{{\sqrt {4{x^2} - 1} }}{x} + C$$

#### Work Step by Step

\eqalign{ & \int {\frac{{\sqrt {4{x^2} - 1} }}{{{x^2}}}dx} \cr & {\text{substitute }}x = \frac{1}{2}sec\theta ,{\text{ }}dx = \frac{1}{2}\sec \theta \tan \theta d\theta \cr & \int {\frac{{\sqrt {4{x^2} - 1} }}{{{x^2}}}dx} = \int {\frac{{\sqrt {4{{\left( {1/2\sec \theta } \right)}^2} - 1} }}{{{{\left( {1/2\sec \theta } \right)}^2}}}\left( {\frac{1}{2}\sec \theta \tan \theta d\theta } \right)} \cr & {\text{simplify}} \cr & = \int {\frac{{4\sqrt {{{\sec }^2}\theta - 1} }}{{{{\sec }^2}\theta }}\left( {\frac{1}{2}\sec \theta \tan \theta d\theta } \right)} \cr & {\text{pythagorean identity}} \cr & = \int {\frac{{2\sqrt {{{\tan }^2}\theta } }}{{{{\sec }^2}\theta }}\left( {\sec \theta \tan \theta d\theta } \right)} \cr & = 2\int {\frac{{{{\tan }^2}\theta }}{{\sec \theta }}d\theta } = 2\int {\frac{{{{\sin }^2}\theta }}{{{{\cos }^2}\theta }}\left( {\cos \theta } \right)d\theta } \cr & = 2\int {\frac{{{{\sin }^2}\theta }}{{\cos \theta }}d\theta } = 2\int {\frac{{1 - {{\cos }^2}\theta }}{{\cos \theta }}d\theta } = 2\int {\left( {\sec \theta - \cos \theta } \right)d\theta } \cr & {\text{integrating}} \cr & = 2\left( {\ln \left| {\sec \theta + \tan \theta } \right| - \sin \theta } \right) + C \cr & {\text{with }}sec\theta = 2x{\text{ and tan}}\theta = \sqrt {4{x^2} - 1} \cr & = 2\left( {\ln \left| {2x + \sqrt {4{x^2} - 1} } \right| - \frac{{\sqrt {4{x^2} - 1} }}{{2x}}} \right) + C \cr & = 2\ln \left| {2x + \sqrt {4{x^2} - 1} } \right| - \frac{{\sqrt {4{x^2} - 1} }}{x} + C \cr}

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